我遇到以下错误，但可以找出原因吗？
MySQL Query Failed: Column 'candidate_ID' in field list is ambiguous
SELECT SQL_CALC_FOUND_ROWS
candidate.candidate_id AS candidateID,
candidate.candidate_id AS exportID,
candidate.is_hot AS isHot,
candidate.date_modified AS dateModifiedSort,
candidate.date_created AS dateCreatedSort,
candidate_ID AS candidateID,
candidate.first_name AS firstName,
candidate.last_name AS lastName,
extra_field0.value AS extra_field_value0,
candidate.city AS city,
candidate.desired_pay AS desiredPay,
candidate.email1 AS email1,
candidate.phone_cell AS phoneCell,
DATE_FORMAT(candidate.date_modified, '%d-%m-%y') AS dateModified,
IF(candidate_joborder_submitted.candidate_joborder_id, 1, 0) AS submitted,
IF(attachment_id, 1, 0) AS **strong text**attachmentPresent
FROM
candidate
LEFT JOIN extra_field AS extra_field0 ON candidate.candidate_id = extra_field0.data_item_id AND extra_field0.field_name = 'Job Title' AND extra_field0.data_item_type = 100
LEFT JOIN attachment
ON candidate.candidate_id = attachment.data_item_id
AND attachment.data_item_type = 100
LEFT JOIN candidate_joborder AS candidate_joborder_submitted
ON candidate_joborder_submitted.candidate_id = candidate.candidate_id
AND candidate_joborder_submitted.status >= 400
AND candidate_joborder_submitted.site_id = 1
AND candidate_joborder_submitted.status != 650 LEFT JOIN saved_list_entry
ON saved_list_entry.data_item_type = 100
AND saved_list_entry.data_item_id = candidate.candidate_id
AND saved_list_entry.site_id = 1
WHERE
candidate.site_id = 1
GROUP BY candidate.candidate_id
ORDER BY dateModifiedSort DESC
LIMIT 0, 15
在我需要的时候提供任何帮助，我们将不胜感激
最佳答案
您的 candidate_ID AS CandidateID， 没有表名称。由于您在两个不同的表中有candidate_ID，因此您必须指定表名称:
candidate.candidate_ID AS candidateID,
这可以避免歧义
关于MySQL查询失败: Column 'candidate_ID' in field list is ambiguous?，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/40270400/